Answer :

Let us draw two intersecting lines, AB and CD.

Let us denote the intersection point of AB and CD as point P.

It is given to us that one of the angles formed by these two intersecting lines, is a right angle, i.e., 90°.

Let us assume ∠CPA is equal to 90°. - - - - (i)

Now, CPD is a straight line and by linear pair axiom,

∠CPA + ∠APD = 180°

⇒ 90° + ∠APD = 180° [From equation (i)]

⇒ ∠APD = 180° - 90°

⇒ ∠APD = 90° - - - - (ii)

Similarly, APB is a straight line and by linear pair axiom,

∠APD + ∠DPB = 180°

⇒ 90° + ∠DPB = 180°

⇒ ∠DPB = 180° - 90°

⇒ ∠DPB = 90° - - - - (iii)

In the same way, ∠CPA + ∠BPC = 180°

⇒ 90° + ∠BPC = 180°

⇒ ∠BPC = 180° - 90°

⇒ ∠BPC = 90° - - - - (iv)

From equations (ii), (iii), and (iv), we find that the rest of the three angles are also 90° each.

Thus, all the four angles are 90° each.

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